Q.The motion of a particle executing simple harmonic motion is described by the displacement function, $x(t) = A\cos(\omega t + \phi)$. If the initial ($t = 0$) position of the particle is 1 cm and its initial velocity is $\omega$ cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi\ \text{s}^{-1}$. If instead of the cosine function, we choose the sine function to describe the SHM: $x = B\sin(\omega t + \alpha)$, what are the amplitude and initial phase of the particle with the above initial conditions.

Concept: Simple Harmonic Motion — the displacement function and its derivative give position and velocity at any instant.

Step 1 — Using the cosine form

Given $x(t) = A\cos(\omega t + \phi)$, with $\omega = \pi\ \text{s}^{-1}$.

At $t = 0$:

$x(0) = A\cos\phi = 1\ \text{cm}$

$v(0) = -\omega A\sin\phi = \omega\ \text{cm/s}$

Step 2 — Solve for $A$ and $\phi$

From the velocity equation: $-\omega A\sin\phi = \omega \Rightarrow A\sin\phi = -1$.

Square and add:

$(A\cos\phi)^2 + (A\sin\phi)^2 = 1^2 + (-1)^2 \Rightarrow A^2 = 2 \Rightarrow A = \sqrt{2}\ \text{cm}$.

Then $\tan\phi = \frac{A\sin\phi}{A\cos\phi} = \frac{-1}{1} = -1 \Rightarrow \phi = -\frac{\pi}{4}$ (since $\cos\phi > 0$, $\sin\phi < 0$).

Step 3 — Using the sine form

$x(t) = B\sin(\omega t + \alpha)$. At $t = 0$:

$x(0) = B\sin\alpha = 1$

$v(0) = \omega B\cos\alpha = \omega \Rightarrow B\cos\alpha = 1$ …

Why this approach works

Simple harmonic motion is fundamentally about a particle oscillating back and forth under a restoring force proportional to displacement. The displacement function $x(t) = A\cos(\omega t + \phi)$ is one standard way to write the solution — here $A$ is the amplitude (maximum displacement), $\omega$ is the angular frequency, and $\phi$ is the initial phase angle (the "head start" in the cosine wave at $t=0$).

The key insight: initial conditions — position and velocity at $t=0$ — completely determine $A$ and $\phi$. You plug $t=0$ into both the displacement and the velocity (which is the derivative of displacement), and solve the two resulting equations. The same logic applies if you choose the sine form $x = B\sin(\omega t + \alpha)$; only the phase interpretation shifts.

Let’s work through it.

Step-by-step solution

1. Write the displacement and velocity for the cosine form

We have:

The velocity is the time derivative:

Given: $\omega = \pi\ \text{s}^{-1}$, initial position $x(0) = 1\ \text{cm}$, initial velocity $v(0) = \omega\ \text{cm/s} = \pi\ \text{cm/s}$.

2. Apply initial conditions to the cosine form

At $t=0$:

(Since $\omega = \pi$, it cancels from both sides of the velocity equation.)

Now we have two equations:

3. Find amplitude $A$

Square and add:

(Amplitude is positive by definition.)

4. Find initial phase $\phi$

From $A\cos\phi = 1$ and $A\sin\phi = -1$, with $A = \sqrt{2}$:

Both conditions together tell us $\phi$ is in the fourth quadrant. The angle whose sine is $-1/\sqrt{2}$ and cosine is $+1/\sqrt{2}$ is:

5. Now switch to the sine form

We now describe the same motion as: …

Step 1: The velocity is the derivative of displacement: for $x(t)=A\cos(\omega t+\phi)$, $v(t)=-A\omega\sin(\omega t+\phi)$.

Step 2: Apply the $t=0$ conditions: $x(0)=A\cos\phi=1$, and $v(0)=-A\omega\sin\phi=\omega \Rightarrow A\sin\phi=-1$.

Step 3: Square and add the two equations: $A^2\cos^2\phi+A^2\sin^2\phi=1^2+(-1)^2 \Rightarrow A^2=2 \Rightarrow A=\sqrt2\ \text{cm}$.

Step 4: $\tan\phi=\dfrac{A\sin\phi}{A\cos\phi}=\dfrac{-1}{1}=-1$; since $\cos\phi>0$ and $\sin\phi<0$, $\phi$ is in the fourth quadrant: $\phi=-\pi/4$.

Step 5: Repeat for $x=B\sin(\omega t+\alpha)$: $x(0)=B\sin\alpha=1$, $v(0)=B\omega\cos\alpha=\omega \Rightarrow B\cos\alpha=1$. …